C12H22O11(s) + 12O2(g)==> 12CO2(g) + 11H2O(g)
You know this is another limiting reagent (LR) problem because amounts are given for BOTH reactants.
mols sugar = grams/molar mass = ?
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols sugar to mols CO2.
Do the same for converting mol O2 to mols CO2.
It is likely that the values you obtain will be different which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the regent producing the smaller value is called the LR.
Now use the smaller value to calculate g CO2. g CO2 = mol CO2 x molar mass CO2
You may want to go back and look at the post down the board for the 200 molecules H2 and 200 molecules O2. I corrected that answer.
When cane sugar reacts with oxygen in living systems, carbon dioxide and water are produced. What weight of carbon dioxide can be produced from the reaction of 15.0 grams of cane sugar with 15.0 grams of oxygen? (Hint: balance the equation)
___C12H22O11(s) + ___O2(g) ___CO2(g) + ___H2O(g)
3 answers
thank you!
its 16.03